Use app×
Ask a Question
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE

A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

← Prev Question Next Question →
0 votes
6.4k views
in Chemistry by (83.8k points)

A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

Challenge Your Friends with Exciting Quiz Games – Click to Play Now!

1 Answer

+1 vote
by (92.5k points)
selected by
 
Best answer

K.E. = P.E.

1/2mv2 = 1/4πεo Ze2/d Hence d = Ze2/4πεo( 1/2mv2)

= 9 x 109 x 79 x ( 1.6 x 10-19)2 /( 1.6 x 10-13 ) [ as 1 MeV = 1.6 x 10-13 J]

= 1.137 x 10-13 m

← Prev Question Next Question →

Find MCQs & Mock Test

  1. JEE Main 2026 Test Series
  2. NEET Test Series
  3. Class 12 Chapterwise MCQ Test
  4. Class 11 Chapterwise Practice Test
  5. Class 10 Chapterwise MCQ Test
  6. Class 9 Chapterwise MCQ Test
  7. Class 8 Chapterwise MCQ Test
  8. Class 7 Chapterwise MCQ Test

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

User Contribution to Sarthaks eConnect
Managed by Sarthaks Digitizing India
...